C++ std::unique_ptr : Why isn't there any size fees with lambdas?

Question

I am reading \"Effective Modern C++\". In the item related to std::unique_ptr it\'s stated that if the custom deleter is a stateless object, then no size fees o

Answer 1

A unique_ptr must always store its deleter. Now, if the deleter is a class type with no state, then the unique_ptr can make use of empty base optimization so that the deleter does not use any additional space.

How exactly this is done differs between implementations. For instance, both libc++ and MSVC store the managed pointer and the deleter in a compressed pair, which automatically gets you empty base optimization if one of the types involved is an empty class.

From the libc++ link above

template  >
class _LIBCPP_TYPE_VIS_ONLY unique_ptr
{
public:
    typedef _Tp element_type;
    typedef _Dp deleter_type;
    typedef typename __pointer_type<_Tp, deleter_type>::type pointer;
private:
    __compressed_pair __ptr_;

libstdc++ stores the two in an std::tuple and some Google searching suggests their tuple implementation employs empty base optimization but I can't find any documentation stating so explicitly.

In any case, this example demonstrates that both libc++ and libstdc++ use EBO to reduce the size of a unique_ptr with an empty deleter.